**Slot -2 – Quantitative Aptitude – Modern Maths – Progressions – Let a(base1), a(base2), …a(base52)**

** Let a(base1), a(base2), … , a(base52) be positive integers such that a(base1) < a(base2) < … < a(base52). Suppose, their arithmetic mean is one less than the arithmetic mean of a(base2), a(base3), …, a(base52). If a(base52) = 100, then the largest possible value of a(base1) is?**

a) 45

b) 20

c) 48

d) 23

**Answer:** d) 23

**Solution: **

We want to maximize the value of a1, subject to the condition that

(a2+ a3 +⋯….+ a52 )/51 -(a1+ a2 +⋯….+ a52 )/52 = 1

Since a(base52) = 100 and all the numbers are positive integers, maximizing a_{1} entails maximizing a_{2}, a_{3} ….a_{51}. The only way to do this is to assume that a_{2}, a_{3}…. A_{52} are in an AP with a common difference of 1.

Let the average of a_{2}, a_{3}, …, a_{52} i.e.

(a2+ a3 +⋯….+ a52 )/51= a_{27} =A ( using the average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term)

So a_{52 } = a_{27} + 25*1 = a_{27} + 25 and

given a_{52 }= 100

=> a_{27} = A = 100 – 25 = 75

a_{2} + a_{3} + … + a_{52 }= 75×51 = 3825

Given a_{1} + a_{2} +… + a_{52 } = 52(A – 1) = 3848

Hence a_{1 }= 3848 – 3825 = 23

Ans : 23

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